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How to find the limit of:

$\lim_{n\to\infty} \sqrt[n]{n+\sin^2n}$

using squeeze theorem?
Because $0\le \sin^2n \le 1 $, I find $a_n=\sqrt[n]{n}$ (which is equal to 1) and $c_n=\sqrt[n]{n+1}$, but I don't know how to prove that second formula is also a 1. Could someone help me solve it please? Thank you!

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  • $\begingroup$ How do you know $\lim \sqrt[n]{n}=n$? You can't just assume. The proof of $\lim \sqrt[n]{n+1}$ isn't much different. $\endgroup$
    – fleablood
    Feb 6, 2020 at 19:32
  • $\begingroup$ If you know that $2^{1/n} \rightarrow 1$ and $n^{1/n} \rightarrow 1,$ then you can use the fact that $\sqrt[n]{n} \leq \sqrt[n]{n + {\sin}^2 n} \leq \sqrt[n]{n+n} = 2^{1/n}n^{1/n},$ and note that the left side approaches $1$ and the right side approaches $1 \cdot 1 = 1.$ $\endgroup$
    – Dave L. Renfro
    Feb 6, 2020 at 19:34
  • $\begingroup$ $n^{1/n} \lt (n+1)^{1/n} \lt (2n)^{1/n}=2^{1/n}n^{1/n}.$ $\endgroup$
    – Peter Szilas
    Feb 6, 2020 at 19:34
  • $\begingroup$ Thank you both! Easier than I thought $\endgroup$
    – L.krez
    Feb 6, 2020 at 19:42

1 Answer 1

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Here's a hint:

$$n<n+1\leq 2n\implies \sqrt[n]{n}<\sqrt[n]{n+1}\leq \sqrt[n]{2n}=\sqrt[n]{2}\cdot\sqrt[n]{n}.$$

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  • $\begingroup$ (+1) Exactly what I gave in a comment, but your answer came first! $\endgroup$
    – Dave L. Renfro
    Feb 6, 2020 at 19:35
  • $\begingroup$ Thank you, now it seems obvious! $\endgroup$
    – L.krez
    Feb 6, 2020 at 19:42

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